A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of comfiguration I?

6/5

3/2

5/3

Mechanics}Small Oscillations

One can derive the frequency of small oscillation for a rigid body in general by using the torque form of Newton's Laws: . (I is moment of inertia, r is moment arm)

In this case, one has a constant downwards force , which acts at a moment arm angle . Thus, , where the approximation works if .

The equation of motion for small angles is thus . This is similar in form to that of a simple harmonic oscillator with the angular frequency being .

Now, the problem is to find the angular frequency for each system.

One needs not worry about the rod, since it is massless and has no moment of inertia.

The moment of inertia of system I is just . The radius of gyration r is just (an r for each mass).

The moment of inertia of system II is . The radius of gyration r is just .

Thus , as in choice (A).

Alternate Solutions

oliTUTilo2012-11-04 12:17:14

I suggest using the Lagrangian formalism with the angle theta as your coordinate and the small angle approximation at hand (you just need for small ). If you're comfortable with it, you can pretty quickly find the frequency of oscillation for the generalized pendulum of system II with pretty much no extraneous memorization. If you can't remember that the simple pendulum's frequency is , you can generalize the distance of the intermediate mass to quickly find the frequencies for both systems.

Let be the distance of the intermediate mass from the pivot.

(Lagrangian)

(Euler Lagrange Eq)

For system I, , the simple pendulum.
For system II,

So the ratio of frequencies of System II and I is

This may have looked long, but in practice you'd hopefully be better experienced with Langrangians and be able to cut corners (like knowing that out of L, only T is going to have in it and only V is going to have in it). Depends on how used to them you are, but you at least don't need to memorize anything you shouldn't have memorized anyways (crucial for me).

BrandeisUDR2016-02-28 00:55:55

The discrepancy in your negative sign before square rooting comes from the fact that the potentials should be positive in the initial Lagrangian, not negative. Otherwise, great alternate solution.

I believe, as always with GRE, there's a way to get to the right answer by a process of elimination knowing only the basic of the formulae.

Simple pendulum: =

Step 1: The answer 's gotta have a square root in it, so eliminate C,D,E

Step 2: Substitute with a simpler problem: both masses at their combined center of mass ( that is instead of one mass being at half length and the second at full length we contemplate a pendulum with both masses at 3/4 length) . This gives us
the final answer has to be less than the latter (because we have to move weights apart , thus adding an additional moment of inertia that slows down the oscillation. Eliminate B.
A. is your only remaining choice.

I suggest using the Lagrangian formalism with the angle theta as your coordinate and the small angle approximation at hand (you just need for small ). If you're comfortable with it, you can pretty quickly find the frequency of oscillation for the generalized pendulum of system II with pretty much no extraneous memorization. If you can't remember that the simple pendulum's frequency is , you can generalize the distance of the intermediate mass to quickly find the frequencies for both systems.

Let be the distance of the intermediate mass from the pivot.

(Lagrangian)

(Euler Lagrange Eq)

For system I, , the simple pendulum.
For system II,

So the ratio of frequencies of System II and I is

This may have looked long, but in practice you'd hopefully be better experienced with Langrangians and be able to cut corners (like knowing that out of L, only T is going to have in it and only V is going to have in it). Depends on how used to them you are, but you at least don't need to memorize anything you shouldn't have memorized anyways (crucial for me).

BrandeisUDR2016-02-28 00:55:55

The discrepancy in your negative sign before square rooting comes from the fact that the potentials should be positive in the initial Lagrangian, not negative. Otherwise, great alternate solution.

I looked at this for awhile, and I know now that saying "radius of gyration" is incorrect, it should be distance from the center of mass. Radius of gyration is defined as . This will get you a value of instead of and instead of . Interestingly both are fairly accurate, that is because is the root mean square value of the distance from the center of mass. If the masses were not equal then your answer would actually turn out to be incorrect. I hope this clarifies thisngs for everyone else. Also I just received a new practice exam of 100 questions a few days ago.

Could anyone tell me why I will be wrong if I go this way: There is no net torque, so L should be conserved. L(f)=L(i) ---> I(f)w(f)=I(i)w(i) ; I(i)=2ml^2 ; I(f)=ml^2+m(l/2)^2, so we will have: w(f)/w(i)=8/5 which is in the choices!!

Setareh2011-11-06 13:30:39

I got my answer, I had a problem in the concept of NET torque.. I thought since the torque is not change in both states so there is no net torque, now I see I was wrong..

This is actually nice n simple no small angle approximations or torque equations needed.

recall the period for physical pendulums:

I is inertial and l is distance to center of mass

invert and times by for angular frequency.

for I and

for II and

plug in and divide to get

62a2016-10-02 23:57:45

Best answer. That\'s one simple and general equation that has certainly been neglected in my education. Small-angle approximations are very useful but should not be emphasized at the expense of exact formulas that are only slightly more complicated!

Once again, Taylor expansion can come to the rescue here. If you know nothing about radii of gyration, you need only remember that any potential well can be approximated by a simple harmonic potential. In this case

We know the potential energy exactly. For situation I, it's . We need only approximate this to second order. Remember . Drop the constant order part to get

Identify . Now we have the familiar formula for a linear oscillator of , but that doesn't apply in an angular oscillator. The mass isn't right. The angular analogue of mass is rotational inertia, . For the first case, , and you get .

Carry this out for case II now. . . Thus, , and . This gets you the ratio 6/5.

= for a pendulum in general, where is the distance from the pivot to the center of mass along the pendulum and is the total mass. Then, for config. I, =. For config. II, =.

After the algebra, you arrive at answer (A).

gliese876d2008-10-08 12:04:22

Thank you for this clarification! I knew for config 1, it should end up being the same as a simple pendulum: but then I was getting confused on the physical pendulum formula and mistakenly using r rather than

gliese876d2008-10-08 12:29:22

Thank you for this clarification! I knew for config 1, it should end up being the same as a simple pendulum: but then I was getting confused on the physical pendulum formula and mistakenly using r rather than

It took me a long time, but I finally figured out what the solution is really saying.

First off, is the general equation. I explicitly show that the length is a moment arm to the center of mass. Next, I point out that "radius of gyration" () is not what the author really means; this is seen in how they calculate it and use it. The radius of gyration is given by , being the total mass. Thus it does not follow superposition as implicitly suggested by the solution, and, in terms of it, the frequency would still depend on : . (Clearly, the radius of gyration is not a useful quantity in this calculation - why add another length? Just use expressed as a fraction of the length of the rod).

So what's really going on? Well, look at their and . Notice that in both cases. Essentially, what they have is , but with the omitted. Their expression for the frequency is . Which, you can see, will lead to a correct answer.

I think a more logical solution would have and represent moment arms to the respective centers of mass, which would have the effect of halving their current values (found by and ). The moments of inertia are correct, so just use the appropriate values for and in the final equation to arrive at the solution.

What is this radius of gyration? Wikipedia doesn't give you much, why can't this be worked with CM? Georgi, its really seeming like your method fails, and its what I tried initially too. Get a CM of 3/4R, get your moment of inertia to be (9/8)MR^2, but then you get , which is fail.

theodiggers2007-10-07 16:17:50

Sorry, I'm an idiot. Can't calculate moments of inertia with CM...

theodiggers2007-10-07 16:27:47

...Freedman and Young gives no reason why this fails in general, only an example of its failure. Why does this fail in general?

The force of gravity acts on the center of mass of each pendulum (not on the end of the rod). System 1 has CM at the end of course:rnrnrnrnif is the length of the rod. System 2 has the CM atrnrn by inspection, or if you like,rnrnrnrnThe used in the official solution should be the moment arm of the center of mass, and is NOT the same for both pendulums.rnrnThe ratio of frequencies is thenrnrnrn

flyboy6212010-10-22 13:35:23

Not sure why my formatting didn't look the same as in the preview. Everywhere you see "rnrn" read as a double carriage return. My point is the "r" used in the official solution is not the same for both pendulums, as it is at the center of mass of each.

I believe that there is a problem with the radii of gyration you have specified. If all the mass were relocated to the radius of gyration (ROG), the moment of inertia would be the same as in the original setup. It doesn't make sense that your ROGs are larger than r, and for configuration I) all the mass already is located at one point and so the ROG should just be r. The corrections I think that need to be made are:

1) In the angular frequency equation, m=2m

2) The ROG in config. I) is just r

3) The ROG in config. II) is 3r/4

This will of course give the same answer.

dirichlet2006-11-27 12:00:48

I think that is not the r of gyration for con 2 either . That would be {sqrt(5/8)}r